I have a following expression for expectation of some random variable $X$:
$E(X(t)|\lambda)=\frac{1}{p}(1-e^{-\lambda pt}) \ $
We assume that $p$ is fixed and $\lambda$ has a following mixing gamma distribution:
$f(\lambda|r,\alpha)=\frac{\alpha^r \lambda^{r-1} e^{-\lambda \alpha}}{\Gamma (r)}, \lambda>0$
I need to develop final formula for $E(X(t)$ - to do this I believe I need to calculate the following:
$E(X(t))=E(X(t)|\lambda) f(\lambda)=\frac{1}{p}-\frac{1}{p} \int_0^\infty e^{-\lambda pt} \frac{\alpha^r \lambda^{r-1} e^{-\lambda \alpha}}{\Gamma (r)} d\lambda $
Can someone please confirm if above is correct? I'm not sure if my reasoning is fine, only final solution without derivation has been presented in the paper, namely: $E(X(t))=\frac{1}{p}-\frac{\alpha ^r}{p(\alpha+pt)^r}$
Thank you
\begin{align}E(X(t))&=\color{red}{\int_0^\infty}E(X(t)|\lambda) f(\lambda)\, \color{red}{d\lambda}\\&=\frac{1}{p} \int_0^\infty (1-e^{-\lambda pt}) \frac{\alpha^r \lambda^{r-1} e^{-\lambda \alpha}}{\Gamma (r)} d\lambda \\ &=\frac1{p} \left( \int_0^\infty \frac{\alpha^r\lambda^{r-1}e^{-\lambda \alpha}}{\Gamma(r)}\, d\lambda - \int_0^\infty \frac{\alpha^r\lambda^{r-1}e^{-\lambda (\alpha+pt)}}{\Gamma(r)}\, d\lambda \right)\\ &=\frac1{p} \left( 1 - \frac{\alpha^r}{(\alpha+pt)^r}\int_0^\infty \frac{(\alpha+pt)^r\lambda^{r-1}e^{-\lambda (\alpha+pt)}}{\Gamma(r)}\, d\lambda \right)\\ &=\frac1{p} \left( 1 - \frac{\alpha^r}{(\alpha+pt)^r}\right)\end{align}
where I have used the property that $\int_0^\infty \frac{\beta^r\lambda^{r-1}e^{-\lambda \beta}}{\Gamma(r)}\, d\lambda =1$ since the function that is being integrated is a density function.